Question

# The divisor when the quotient, dividend and the remainder are respectively 547, 171282 and 71 is equal to A.333B.323C.313D.303

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Hint: First we will first assume that $x$ be the divisor and then substitute the values of remainder, dividend and quotient in the formula, ${\text{Dividend}} = {\text{Divisor}} \cdot {\text{Quotient + Remainder}}$ to find the required divisor.

We are given that the divisor when the quotient, dividend, and the remainder are respectively 547, 171282, and 71.

Let us assume that $x$ be the divisor.

Using the formula, ${\text{Dividend}} = {\text{Divisor}} \cdot {\text{Quotient + Remainder}}$ in the given values, we get
$\Rightarrow 171282 = x \times 547 + 71 \\ \Rightarrow 171282 = 547x + 71 \\$

Subtracting the above equation by 71 on both sides, we get
$\Rightarrow 171282 - 71 = 547x + 71 - 71 \\ \Rightarrow 171211 = 547x \\$

Dividing the above equation by 547 on both sides, we get
$\Rightarrow \dfrac{{171211}}{{547}} = \dfrac{{547x}}{{547}} \\ \Rightarrow 313 = x \\ \Rightarrow x = 313 \\$

Thus, the required divisor is 313.
Hence, option C is correct.

Note: In solving these types of questions, students should remember that the role of the division is to break the number into equal parts, so the number, which divides another number is divisor whereas the number which is being divided is termed as a dividend, so one should know the difference.